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I recently worked on the following idea: Eigenvalue of an Euler product type operator?

Summary of the idea

We represent numbers by infinite dimensional matrices such as $3$ will have all $0$s except the $1$st row will have a $1$ in the third column, the $2$nd row will have a $1$ in the $6$th row and the $r$th row and $3r$th column also have $1$s:

$$ \hat 3 = | 1 \rangle \langle 3 | + | 2 \rangle \langle 6 | + | 3 \rangle \langle 9 | + \dots = \begin{bmatrix} 0 &0 & 1 &0 & 0& \dots & 0 \\ 0 &0 & 0 &0 &0 & 1 & \dots & \\ \vdots \\ 0 & 0 & 0 & \dots \end{bmatrix} $$

Similarly we can define $\hat 2$:

$$ \hat 2 = | 1 \rangle \langle 2 | + | 2 \rangle \langle 4 | + | 3 \rangle \langle 6 | + \dots =\begin{bmatrix} 0 &1 & 0 &0 & 0& \dots & 0 \\ 0 &0 & 0 &1 &0 & 0 & \dots & \\ 0 &0 & 0 &0 &0 & 1 & \dots & \\ \vdots \end{bmatrix} $$

One notices that these numbers obey multiplication

$$ \hat 6 = \hat 3 . \hat 2 = \hat 2 . \hat 3$$

where the $r$th row of the $6r$th column has a $1$

One can also use this to define an Euler like product formula:

$$ \hat \zeta (s) = \hat 1 + \hat 2^s + \hat 3^s + \dots = (1- \hat 2^s)^{-1}(1- \hat 3^s)^{-1}(1- \hat 5^s)^{-1} \dots$$

Note: $\zeta(1) |\lambda \rangle = |\text{factors of } \lambda \rangle$

My Observation

Let us define the following ladder operators $a$ and $a^\dagger$ from quantum mechanics:

$$ A^\dagger|n \rangle = | n+1 \rangle$$

Now we make the following observation:

$$ \frac{A^\dagger}{\hat I- A^\dagger} \geq \hat \zeta(1) - \hat I$$

Where I is the identity or $\hat 1$. To see what $\frac{A^\dagger}{1- A^\dagger} $ looks like apply it $\hat 1$ as: $ \frac{A^\dagger}{1- A^\dagger} \hat 1$. In the sense:

$$ \langle m |\frac{A^\dagger}{\hat I- A^\dagger} |n \rangle \geq \langle m |(\hat \zeta(1) - \hat I) |n \rangle $$

In fact for $n>1$:

$$ \langle m |\frac{A^{\dagger 2}}{\hat I- A^\dagger} |n \rangle \geq \langle m |(\hat \zeta(1) - \hat I) |n -1 \rangle $$

Subtracting the above equations:

$$ \langle m|A^\dagger| n \rangle \geq \langle m |(\hat \zeta(1) - \hat I) ( |n \rangle - |n -1 \rangle) \geq \langle m |(\hat \zeta(1) - \hat I) |n \rangle $$

Question

Given a Hamiltonian can be expressed as annihilation and creation operators can the above expression be used to get some lower bounds on the difficult to compute spectrums?

$$H(a^\dagger, a) \geq H'(\hat \zeta,\hat \zeta^\dagger) $$

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  • $\begingroup$ Your multiplicative operators are all lowering, annihilation type operators, not raising, creation ones. $\endgroup$ Dec 26, 2018 at 21:49

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